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#11
October 2nd, 2011, 09:12 PM
 Machinist Executive Member Join Date: Sep 2009 Location: Here n there......midwest Posts: 985
OH no!!!

I'm not getting in the middle of this.....
Heck i might even learn something............

Machinist

Last edited by Machinist; October 2nd, 2011 at 09:24 PM.
#12
October 2nd, 2011, 10:58 PM
 sagefr0g Executive Member Join Date: Apr 2006 Posts: 5,141

Quote:
 Originally Posted by Machinist I'm not getting in the middle of this..... Heck i might even learn something............ Machinist
hmmm, well Macho, as i alluded to earlier, the problem is really more complicated in that the cost of the ride varies as the number of competitors join in the fray, no?
errrhh, uhhmm, ok, lol, i think for every equal competitor that joins in the carnival reduces the cost by a factor of one. so two riders, the cost is halved, twelve riders the cost is cut by one twelfth, no?
so if that's the case i think the EV = \$29/12 = ~ \$2.42
cause using tthree's equation EV = (Y - 12*X)/12 for no reduced cost becomes EV = (Y - 12*X/12)/12 = (Y-X)/12 taking into account the reduced cost...... & Y - X = \$29 as tthree stipulated or Y = \$29 + X, so substituting for Y, we have EV = (\$29 + X - X)/12 or EV = \$29/12 = ~\$2.42
no?
sorry for all the confusion.
is this then properly evaluated assuming the cost is reduced as stipulated?
#13
October 3rd, 2011, 06:36 AM
 London Colin Senior Member Join Date: Aug 2009 Posts: 378

Sage, I get the impression that there is miscommunication going on here on a massive scale. The only way anyone is going to know with certainty if they have understood you is if you post an entire example with real numbers, and no algebra whatsoever.

It's still not clear to me from what you've written, whether each rider pays the same amount or whether there is some variation among them.

In general, the simplest way to think about any competition is this -

After you pay to enter, the EV is the total prize pool (which might, as in this case, be a single prize) divided by the number of entrants.

Before you pay, the EV is the above figure, minus the amount you would have to pay. (And this tells you whether it is worthwhile to participate)

If you don't know in advance what you will have to pay, or how many competitors you will face, then you need to estimate average values.
#14
October 3rd, 2011, 07:50 AM
 Machinist Executive Member Join Date: Sep 2009 Location: Here n there......midwest Posts: 985

Now ya did it Sagefr0g,
You half confused everybody and totally confused yourself. Now ya painted yourself into a corner, now what, ?????? I'd wait for the paint to dry, of course i dont think i would paint my way into the corner in the first place. You might want to hop over to Y and have her wipe the paint off your feet.........

Go ahead froggy , give us some real numbers and see what ya come up with....... Like Collin said your confusing the hell out of us all!!!!!!
I hate to see my Froggy friend crosseyed all day just sitting on a lillypad staring up at the blue sky trying to get this situation understood.
This might be interesting....
Use easy numbers though frog like 10, and 100's , easily divisible numbers as small as possible......

Machinist
#15
October 3rd, 2011, 11:27 AM
 blackriver Senior Member Join Date: Dec 2010 Posts: 364

I'm pretty sure you are correct if what you said its what you mean. It sounds like they charge the same thing every time and they divide that cost by the number of riders. There is also 1 prize and it is effectively divided (ev wise) among will the riders. So the difference between the price they charge to run it minus the prize value is always the same and you should just divide that by the number of riders
#16
October 3rd, 2011, 12:36 PM
 sagefr0g Executive Member Join Date: Apr 2006 Posts: 5,141

Quote:
 Originally Posted by blackriver I'm pretty sure you are correct if what you said its what you mean. It sounds like they charge the same thing every time and they divide that cost by the number of riders. There is also 1 prize and it is effectively divided (ev wise) among will the riders. So the difference between the price they charge to run it minus the prize value is always the same and you should just divide that by the number of riders
there yah go, i think you pretty much summed it up........

it was my bad to poorly word the scenario, then up and change the scenario in the midst of the whole thread. i apologize for that, everybody.
so but, thank you everyone very much for your input, as i think i genuinely now understand the situation better.........
and sorry for being so obtuse, but i'm sure we all understand why that's necessary.
i'll try and sum up the situation and lay some hard numbers on it as Londin Colin suggested.
#17
October 3rd, 2011, 01:37 PM
 Richard Munchkin Senior Member Join Date: Jul 2011 Location: California Posts: 145

Sage,

Just to sum up why it it important.

If you pay 1 cent to ride and the prize is \$29.01 then you obviously have positive ev with 12 riders.

If you pay \$100 to ride and the prize is \$129 then you would have big negative ev with 12 riders. (but your ev would be \$29 for one rider.)
#18
October 3rd, 2011, 01:54 PM
 London Colin Senior Member Join Date: Aug 2009 Posts: 378

Quote:
 Originally Posted by blackriver I'm pretty sure you are correct if what you said its what you mean. It sounds like they charge the same thing every time and they divide that cost by the number of riders. There is also 1 prize and it is effectively divided (ev wise) among will the riders. So the difference between the price they charge to run it minus the prize value is always the same and you should just divide that by the number of riders
Ah, that makes sense. And it's pretty easy to prove that Sage has it right -

Suppose the total charged is \$12, and thus the total given out is \$41 to give the \$29 net win for a single rider, as described. A lone rider is charged \$12 to compete, whereas he is only charged \$1 on a full ride of 12 competitors.

So for different numbers of players we get EVs of

1: \$41 - \$12 = \$29 (same as \$29/1)
2: \$41/2 - \$6 = \$14.50 (same as \$29/2)
.
.
12: \$41/12 - \$1 = \$2.41667 (same as \$29/12)

And going back to the algebra, the above all works out because -

V/N -C/N = (V-C)/N

where V = Value (e.g., \$41), C = Cost (e.g., \$12), and N = Number of players.

QED
#19
October 3rd, 2011, 09:59 PM
 Nynefingers Senior Member Join Date: Oct 2009 Posts: 355

Sage,

It might be easier just to back up and look at it from a general view. Back to the EV basics.

EV = (chance of winning)*(amount won) - (chance of losing)*(amount lost)

Since the amount won is the net win after you pay the cost of entry into the game, this can be simplified to:

EV = (chance of winning)*(prize) - (entry cost)

It sounds like in this case the amount of the prize is constant (at least for the sake of argument it is...we know the Expected Value of the prize) and the cost to enter depends on the number of people. Obviously the chance of winning also depends on the number of people.

Using the numbers London Colin chose, which is a \$41 prize and a total cost of \$12 split evenly between all players, we get the following:

1 player: (1/1)*(\$41)-(\$12/1) = \$29
2 players: (1/2)*(\$41)-(\$12/2) = \$14.50 = \$29/2
3 players: (1/3)*(\$41)-(\$12/3) = \$9.67 = \$29/3
.
.
.
12 player: (1/12)*(\$41)-(\$12/12) = \$2.42 = \$29/12

This is of course exactly what London Colin stated.

For a situation where your entry fee is constant regardless of how many people are entered (I'll use a \$34 prize and a \$5 entry cost), you get this:

1 player: (1/1)*(\$34)-(\$5) = \$29
2 players (1/2)*(\$34)-(\$5) = \$12
3 players (1/3)*(\$34)-(\$5) = \$6.33
4 players (1/4)*(\$34)-(\$5) = \$3.50
5 players (1/5)*(\$34)-(\$5) = \$1.80
6 players (1/6)*(\$34)-(\$5) = \$0.67
7 players (1/7)*(\$34)-(\$5) = -\$0.14
8 players (1/8)*(\$34)-(\$5) = -\$0.75
9 players (1/9)*(\$34)-(\$5) = -\$1.22
10 players (1/10)*(\$34)-(\$5) = -\$1.60
11 players (1/11)*(\$34)-(\$5) = -\$1.91
12 players (1/12)*(\$34)-(\$5) = -\$2.17

This is what Richard Munchkin was describing.

Hopefully this all makes sense to you and I didn't just make it worse. If not, you can send me an email with some actual numbers and maybe I can explain it better. You don't have to say what game it is if you don't want to, but I think some real numbers would make it easier to explain to you.
#20
October 4th, 2011, 12:33 AM
 sagefr0g Executive Member Join Date: Apr 2006 Posts: 5,141
tthree, blackriver, London Colin, Nynefingers, i get your drift

thank you again all, and sorry again for the poorly worded model, especially sorry for the last minute stipulation about the cost decreasing as other riders entered the fray.
but anyway, both scenarios are i think interesting, ie. the constant cost scenario and the decreasing cost scenario.

here's my final take on both scenarios at this point (i pretty much relied on tthree's equations:

there is a merry go round with twelve horses and a brass ring that has value greater than the cost of the ride.
a lone rider will always experience plus ev, cause he is gonna get the brass ring.
but if the ride fills up with more riders and the value of the brass ring and the cost of the ride doesn't change, then there may reach a point where our hero will experience either zero ev or negative ev if the numeric value of the ratio of the value of the brass ring to the cost of the ride isn't high enough
.

that's dependent upon the following:
Y = value of the prize
x = cost of the ride
where Y>X
P = number of riders
then
EV = (Y - P*X)/P
or
EV = [Y - X - (P-1)*X]/P
if P = 1 then ev is always positive
but if (p-1)*X >= (Y - X) then ev <= 0
alternatively
if P - 1 >= Y/X - 1 then ev <= 0
and
if P >= Y/X then ev <= 0

so if Y/X > P then ev > 0

so if Y/X > 1 then ev > 0 for one rider (which was a given)
if Y/X > 2 then ev > 0 for two riders
if Y/X > 3 then ev > 0 for three riders
if Y/X > 4 then ev > 0 for four riders
if Y/X > 5 then ev > 0 for five riders
if Y/X > 6 then ev > 0 for six riders
if Y/X > 7 then ev > 0 for seven riders
if Y/X > 8 then ev > 0 for eight riders
if Y/X > 9 then ev > 0 for nine riders
if Y/X > 10 then ev > 0 for ten riders
if Y/X > 11 then ev > 0 for eleven riders
if Y/X > 12 then ev > 0 for twelve riders

so the ratio of the brass ring value to ride cost must have a greater numerical value than than actual number of riders inorder for play to be plus ev.

but suppose that the carnival had a promotion where the cost X of the ride was reduced by dividing the original ride cost by the number of riders while keeping the value of the brass ring the same.
now ev > 0 irrelevant of the number of riders.
this is true because to begin with Y/X > 1 and dividing X by any number two through twelve is going to result in some number greater than one being multiplied by those numbers so the ratio Y/X will end up greater than the number of riders, hence plus ev.

well, whatever, this all is just a model, that roughly, very roughly approximates a real life situation that can be taken advantage of, understanding of the model helps clarify some questions and concerns i had.

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